1.72, Groundwat er Hydrology Prof. Charles Harvey
Le ct u r e Pa ck e t # 4 : Con t inu it y a n d Flow N e t s
Equa t ion of Con t inu it y
• Our equat ions of hydrogeology are a com binat ion of
o Conservat ion of m ass
o Som e em pir ical law ( Darcy’s Law , Fick’s Law )
• Develop a cont rol volum e, rect angular parallelepiped, REV ( Represent at ive Elem ent ary Volum e)
z
y
x dx
dy dz
x, y, z
q
m ass inflow rat e – m ass out flow rat e = change in m ass st orage qx = specific discharge in x- direct ion ( volum e flux per area)
at a point x,y,z L3/ L2- T
Consider m ass flow t hrough plane y- z at ( x,y,z)
x ρ dy dz L/ T M/ L3 L L = M/ T
Rat e of change of m ass flux in t he x- direct ion per unit t im e per cross- sect ion is
∂
]
[
ρ
q
dydz
dx
dy dz
x, y, z x
∂
x
m ass flow int o t he ent ry plane y- z is
∂
dx
]
[
ρ
q
xdydz
−
[
ρ
q
x]
dydz
∂
x
2
And m ass flow out of t he exit plane y- z is
∂
dx
]
[
ρ
q
xdydz
+
[
ρ
q
x]
dydz
∂
x
2
I n t he x- direct ion, t he flow in m inus t he flow out is
∂
]
−
[
ρ
q
xdxdydz
∂
x
Sim ilar ly, t he flow in t he y- direct ion t hrough t he plane dxdz ( figure on left )
dy dz
x, y, z
dx
dy dz
x, y, z
dx
⎡
⎤ ⎡
⎤
∂
⎢
[
ρ
q
dxdz
−
∂
[
ρ
q
y]
dy
dxdz
⎥ − ⎢
[
ρ
q
dxdz
+
∂
[
ρ
q
y]
dy
dxdz
⎥
=
−
∂
y
[
ρ
q
dxdydz
⎣
y]
∂
y
2
⎦ ⎣
y]
∂
y
2
⎦
y]
for t he net y- m ass flux.
Sim ilarly, we get for t he net m ass flux in t he z- direct ion:
∂
]
−
[
ρ
q
zdxdydz
∂
z
The t ot a l m a ss flu x ( flow out of t he box) is
⎡ ∂
[
ρ
q
]
∂
[
ρ
q
y]
−
∂
[
ρ
q
]
⎤
⎥
dxdydz
x
−
z⎢−
x
St eady- st at e flow equat ion for het erogeneous, anisot ropic condit ions:
⎡
⎞
⎤
For isot ropic, hom ogeneous condit ions ( K is not direct ional)
0
Note that the full equation at this point can be written in summation notation as:
ij
)
0
0
Flow N e t s
As w e have seen, t o w ork w it h t he groundw at er flow equat ion in any m eaningful w ay, w e have t o find som e kind of a solut ion t o t he equat ion. This solut ion is based on boundary condit ions, and in t he t ransient case, on init ial condit ions.
Let us look at t he t w o- dim ensional, st eady- st at e case. I n ot her words, let t he follow ing equat ion apply:
∂
⎛
∂
x
⎝
⎜⎜
K
∂
∂
h
y
y
⎞
∂
∂ ⎞
⎟ +
∂
h
x
⎛
⎜
=
∂
x
⎝
K
x⎠
⎟⎟
⎠
( Map View)A solut ion t o t his equat ion requires us t o specify boundary condit ion. For our purposes w it h flow net s, let us consider
∂
∂
boundary) .
• Const ant - head boundaries (h = const ant )
• Wat er- t able boundary ( free surface, h is not a const ant )
A relat ively st raight forward graphical t echnique can be used t o find t he solut ion t o t he GW flow equat ion for m any such sit uat ions. This t echnique involves t he
h
n
const ruct ion of a flow n e t.
A flow net is t he set of equipot ent ial lines ( const ant head) and t he associat ed flow lines ( lines along w hich groundw at er m oves) for a part icular set of boundary condit ions.
• For a given GW flow equat ion and a given value of K, t he boundary condit ions com plet ely det erm ine t he solut ion, and t herefore a flow net .
I n addit ion, let us first consider only hom ogeneous, isot ropic condit ions:
0
No- flow boundaries (
=
, where n is t he direct ion perpendicular t o t he•
∂
2h
∂
2h
=
+
( Cross- Sect ion)x
∂
2z
C D
gr a ve l h1
E
1
5 2 4
3 7 8
h
h2
6 Sa n d a n d
I m pe r m e a ble La ye r
D a m
F
O G
A B
x y
h h1
C D
G
E F
B A
M y
h2
C h = 0
A h = h
h = h h2
H
B 6 0 ’
5 0 ’ 1 6 ’
G
q= q q5
q4
q3
q2
q1
q= 0 D
Let ’s look at flow in t he vicinit y of each of t hese boundaries. ( I sot ropic, hom ogeneous condit ions) .
N o- Flow Bou n da r ie s:
∂
= 0
∂
h
x
or∂
= 0
∂
h
y
or∂
= 0
∂
h
n
• Flow is parallel t o t he boundary.
• Equipot ent ials are perpendicular t o t he boundary Con st a n t - H e a d Bou nda r ie s: h = const ant
• Flow is perpendicular t o t he boundary.
• Equipot ent ials are parallel t o t he boundary. W a t e r Ta ble Bou n da r ie s: h= z
Anyw here in an aquifer, t ot al head is pressure head plus elevat ion head: h = ψ + z
However, at t he wat er t able, ψ = 0. Therefore, h = z
Ru le s for Flow N e t s ( I sot r opic, H om oge ne ou s Syst e m ) :
I n addit ion t o t he boundary condit ions t he follow ing rules m ust apply in a flow net : 1) Flow is perpendicular t o equipot ent ials everywhere.
2) Flow lines never int ersect .
3) The areas bet ween flow lines and equipot ent ials are “ curvilinear squares” . I n ot her words, t he cent ral dim ensions of t he “ squares” are t he sam e ( but t he flow lines or equipot ent ials can curve) .
• I f you draw a circle inside t he curvilinear square, it is t angent ial t o all four sides at som e point .
1
2 D a m
h= h
h= h Re se r voir
Why are t hese circles? I t preserves dQ along any st ream t ube. dQ = K dm ; dh/ ds = K dh
ds dQ
dQ
dQ dm
I f dm = ds ( i.e. ellipse, not circle) , t hen a const ant fact or is used. Ot h e r point s:
I t is not necessary t hat flow net s have finit e boundaries on all sides; r egions of flow t hat ext end t o infinit y in one or m ore direct ions are possible ( e.g., see t he figure above) .
A flow net can have “ part ial” st ream t ubes along t he edge. A flow net can have part ial squares at t he edges or ends of t he flow syst em .
Ca lcu la t ion s fr om Flow N e t s:
Probably t he m ost im port ant calculat ion is discharge from t he syst em . For a syst em wit h one recharge area and one discharge area, we can calculat e t he discharge wit h t he following expression:
Q = nfK dH H = nd dH Gives: Q = nf/ nd KH
Where Q is t he volum e discharge rat e per unit t hickness of sect ion perpendicular t o t he flow net ; nf is t he num ber of st ream t ubes ( or flow channels) ; nd is t he num ber
For hom ogeneous K,
∂
2I nt roduce t he t ransform ed variable
Applying t his variable gives:
Kx is t he hydraulic conduct ivit y horizont ally on your page, and Kz is t he hydraulic conduct ivit y vert ically on your page. This t ransform at ion is not specific t o t he x-dim ension or t he y- x-dim ension.
2. On t he t ransform ed syst em , follow t he exact sam e principles for flow net s as out lined for a hom ogeneous, isot ropic syst em .
3. Perform t he inverse t ransform on t he syst em , i.e.
K
zZ
=
Z
'
K
x4. I f any flow calculat ions are needed, do t hese calculat ions on t he
hom ogeneous ( st ep 2) sect ion. Use t he follow ing for hydraulic conduct ivit y:
K
'
=
K
xK
zWhere K’ is t he hom ogeneous hydraulic conduct ivit y of t he t ransform ed sect ion. ( NOTE: This t ransform ed K’ is not real! I t is only used for calculat ions on t he t ransform ed sect ion.)
Ex a m ple s:
T I
h = 0
1 layer syst em , you w ill only have curvilinear squares in one of t he layers. Which layer t o draw squares in is your choice: in general you should choose t he t hicker/ larger layer.
2
You can rearrange t he t angent law in any way t o det erm ine one unknown quant it y. For exam ple, t o det erm ine t he angle θ2:
⎛
⎜⎜
K
K
⎞
θ
2=
tan
−1
tan
⎟⎟
1
⎠
2
θ
1
⎝
One im port ant consequence for a m edium wit h large cont rast s in K: high- K layers w ill oft en have alm ost horizont al flow ( in general) , w hile low - K layers w ill oft en have alm ost vert ical flow ( in general) .
Ex a m ple :
I n a t hree- layer syst em , K1 = 1 x 10- 3 m / s and K2 = 1 x 10- 4 m / s. K3 = K1. Flow in t he syst em is 14o below horizont al. What do flow in layers 2 and 3 look like?
K2
K3 K1 76o
-4
θ
=
tan
−1⎛
⎜⎜
1x10
tan
76
o⎟⎟
⎞
=
22
o⎝
1x10
-3⎠
When draw ing flow net s w it h different layers, a very helpful quest ion t o ask is “ What layer allows wat er t o go from t he ent rance point t o t he exit point t he easiest ?” Or, in ot her words, “ What is t he easiest ( frict ionally speaking) way for wat er t o go from here t o t here?”
K2 K1
K1 K2
K2 K1
K
K
1
